Physics- Energy problem?

1. The problem statement, all variables and given/known data
Two baseballs are thrown off the top of a building that is 7.32 m high. Both are thrown with initial speed of 58.7 mph. Ball 1 is thrown horizontally, and ball 2 is thrown straight down. What is the difference in the speeds of the two balls when they touch the ground? (Neglect air resistance.)

vball 1 - vball 2 = ?

2. Relevant equations

Ei = Ef

3. The attempt at a solution

58.7 mph -> 26.08 m/s ? (not sure)

I used

Ei= Ef

Ball 1 ( x - direction)

Ki (0) + Ui (mgh)= 1/2mv^2+ Uf(0)

gh= 1/2 V^2

V= 11.97m/s ( same velocity for Ball 2 (y-direction))

I just dont know how to find the difference. If someone can explain it in steps. Thanks

Yep. Since the question is asking for the difference in speed, as Ball 1 falls to the ground it gains vertical velocity as well (how else would it fall?), and Ball 2 will fall to the ground, gaining vertical velocity.

Let's keep the horizontal velocity of both balls out of the picture for now.

The vertical velocity of Ball 1 will increase from 0 to X as it falls to the ground,

whilst the vertical velocity of Ball 2 will increase from 26.24(thrown to the floor) to (26.24 + Y) as it falls to the ground. Calculating it is simply, v^2 = u^2 + 2AD where A is acceleration due to free fall and D is the height of the building.

After which, we look at horizontal velocity.

Since the horiz velocity of ball 2 is 0, and will likely stay that way.

The horiz of Ball 1 is constant, which is 26.241.

Hence, just using Pythagoras' on Ball 1, you can find the speed for Ball 1.